1. Chapter 14 Chemical Equilibrium 14.2–14.3
The Concept of Dynamic Equilibrium & The Equilibrium Constant (K)

2. 14.2) The Concept of Dynamic Equilibrium
What is dynamic equilibrium?
Dynamic Equilibrium for a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction.
*Recall:
The reaction rates generally increase with increasing concentration of the reactants (unless the reaction order is zero) and decrease with the decreasing concentration of the reactants

3. SO LET’S CONSIDER THE REACTION BETWEEN HYDROGEN & IODINE:
H2(g) + l2(g) ⇌ 2HI(g)
A reaction such as this one can proceed in both the forward and reverse direction and therefore is said to be “reversible”

4. At this point dynamic equilibrium is reached
H2(g) + l2(g) ⇌ 2HI(g)
So let’s say we begin with H2 and I2 in a container. What happens? Initially H2 and I2 begin to react which in turn decreases the rate of the forward reaction. HI begins to form and its concentration begins to increase
The reverse reaction begins to occur at a higher and faster rate. Eventually the rate of the reverse reaction (which has been increasing) equals the forward reaction (which has been decreasing)
At this point dynamic equilibrium is reached

5. WE CAN OBSERVE THIS BY LOOKING AT THIS DIAGRAM:
Note:
Just because the concentrations of reactants and products no longer change at equilibrium does not mean that the concentrations of reactants and products are equal to one another at equilibrium!

6. IMPORTANT:
In Dynamic equilibrium the forward and reverse reactions are still occuring; however, they are occuring at the same rate.
So once dynamic equilibrium is reached the concentrations for H2 and I2, and HI no longer change
NEED TO KNOW!
*** DYNAMIC EQUILIBRIUM ONLY OCCURS WITH GASES AND AQUEOUS SOLUTIONS***

7. DYNAMIC EQUILIBRIUM: AN ANALOGY
Population of country A goes down as the population of country B goes up. As people leave country A, however, the rate in which they leave slows down, because as country A becomes less populated, the pool of potential emigrants gets smaller
On the other hand, as people move into country B, it gets more crowded and some people begin to move from country B to country A

8. DYNAMIC EQUILIBRIUM: AN ANALOGY
As the population of Country B continues to grow, the rate of people moving out of country B gets faster, and eventually the rate of people moving out of country A equals the rate of people moving out of Country B
This is when Dynamic Equilibrium is reached
When dynamic equilibrium is reached, their populations no longer change. Also, just because the rates are the same does not necessarily mean that the populations are the same

9. 14.3 The Equilibrium Constant (K)
The Equilibrium constant is a way to quantify the concentration of the reactants and products at equilibrium

10. Let’s consider the following general chemical equation:
aA + bB ⇌ cC + dB
A and B are the reactants
C and D are the products
a,b,c,d are the respective stoichiometric coefficients

11. aA + bB ⇌ cC + dB
The equilibrium constant (K) for the reaction is defined as the ratio (at equilibrium) of the concentrations of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their stoichiometric coefficients:
Law of Mass Action:

12. Expressing Equilibrium Constants for Chemical Reactions
Let’s consider the following chemical equation:
2N2O5(g) ⇌ 4NO2(g) + O2 (g)
The equilibrium would be expressed by applying the Law of Mass Action:

13. CH4(g) + 2H2S (g) ⇌ CS2(g) + 4H2(g)
Practice Problem:
Express the equilibrium constant for the following chemical equation:
CH4(g) + 2H2S (g) ⇌ CS2(g) + 4H2(g)
[CS2][H2]4
[CH4][H2S]2

14. The Significance of the Equilibrium Constants
Now that you know how to express the Equilibrium constant, but what does it mean?
Large equilibrium constant (K>>1) indicates that the amount of product at equilibrium is larger than the amount of reactant at equilibrium. Therefore, when the equilibrium constant is large, the forward reaction is favored.
For example consider this chemical equation:
H2(g) + Br2(g) ⇌ 2HBr(g) K= 1.9 x 1019 (at 25°C)
This equilibrium constant is large, indicating that the equilibrium point for the reaction lies far to the right – high concentration of the products, low concentration of the reactants –

15. *Remember: the equilibrium constant says nothing about how fast a reaction reaches equilibrium, only how far the reaction has proceeded once equilibrium is reached
A Reaction with a large K value may be kinetically very slow and take a long time to reach equilibrium
On the other hand, a small equilibrium constant (k<< 1) indicates that the reverse reaction is favored and that there will be more reactant than product when equilibrium is reached

16. TO SUMMARIZE:
K<< 1 reverse reaction is favored; forwards reaction does not proceed very far
K ≈ neither direction is favored; forward reaction proceeds about half way
K>> 1 forward reaction is favored; forward reaction proceeds closer to completion

17. PRACTICE PROBLEM: EQUILIBRIUM CONSTANT
The equilibrium constant for the reaction A(g) ⇌ B(g) is 10. A reaction mixture initially contains [A] = 1.1 M and [B] = 0.0 M. which statement is true at equilibrium?
The reaction mixture will contain [A] = 1.0 M and [B] = 0.1 M
The reaction mixture will contain [A] = 0.1 M and [B] =1.0 M
The reaction mixture will contain equal concentrations of A and B
The answer is b) because when applying the Law of Mass Action the products are divided by the reactants so by placing [1.0]/[0.1] = 10. Ignore the initial values given.

18. Relationship between the Equilibrium Constant and the Chemical Equation
If a chemical equation is modified in some way, then the equilibrium constant for that equation changes because of the modification. There are three modifications; they are all fairly similar

19. If you reverse the equation, invert the equilibrium constant.
For example consider this equation:
A + 2B ⇌ 3C
The expression for the equilibrium constant is:
If we reverse this reaction: 3C ⇌ A + 2B
Then according to the law of mass action the equilibrium constant is expressed as:

20. 2. If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to the same factor.
Consider the same equation: A + 2B ⇌ 3C
If we multiply the equation by n, we get: nA + n2B ⇌ n3C
Applying the law of mass action, the expression for the equilibrium constant becomes:

21. 3. If you add two or more individual equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain the overall equilibrium constant.
Consider these two chemical equations and their corresponding equilibrium constant expressions:

22. The two equations sum as follows:
According to the law of mass action, the equilibrium constant for this overall equation is then:

23. Notice that Koverall is the product of K1 and K2 :

24. Practice Problem #1: The Equilibrium Constant and the Chemical Equation
The reaction A(g) ⇌ 2B(g) has an equilibrium constant of K = What is the equilibrium constant for the reaction B(g) ⇌ 1/2A(g)? 1 10
100
0.0010
The answer is 10 because the expression for K for the first chemical equation is k = [B]2/ [A]; therefore, the K is expressed as K = [A]1/2/[B] for the second reaction. So by applying modification rule #2 the expression is written as: (1/ 0.010)1/2 = 10

25. Practice Problem #2: Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation
Consider the following chemical equation and equilibrium constant for the synthesis of ammonia at 25°C:
N2(g) + 3H2(g) ⇌ 2NH3(g) K = 5.6 x 10 5
Calculate the equilibrium constant for the following at 25°C:
NH3(g) ⇌ 1/2N2(g) + 3/2H2(g) K’ = ?

26. Answer to Practice Problem #2:
Begin by reversing the given reaction and taking the inverse value of K:
N2(g) + 3H2(g) ⇌ 2NH3(g) K = 5.6 x 105
2NH3(g) ⇌ N2(g) + 3H2(g) Kreverse = 1 / 5.6 x 105
Next, Multiply the reaction by ½ and raise the equilibrium constant by the ½ power:
NH3(g) ⇌ ½ N2(g) + 3/2H2(g)
K’ = K1/2reverse = (1/ 5.6 x 105)1/2
K = 1.3 x 10-3